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Yesterday I found what I thought to be the typical "99% of readers won't be able to solve it" image, with a simple math question. Despite its looks, it yielded this polynomic equation with the imaginary unit as the exponent.

$$b^{\pm i}-b = 3.$$

The $\pm$ symbol is due to the exponent being the solution of a previous equation, which had two different solutions.

I've been able to solve it numerically by using Halley's method, but I'm struggling to find the exact solutions. By using some complex number properties I've been able to change the base to $e$, turn everything into polar form and separate each equation into a system, equaling the real part to $3$ and the imaginary part. Still, I don't know how to continue. This is what I've done so far.

My first objective was to have an equation system working on the $\mathbb{R}$ domain. To do so, I change the base to $e$, so I can use Euler's formula later.

$$e^{\pm i \log(b)} = 3+b.$$

Then I apply the definition of the complex logarithm that states $\log(x) = \log(|x|)+i \cdot \arg(x)$. This definition may be wrong since I don't remember very well the stuff about the complex logarithm. I do this to prevent having $\sin$ and $\cos$ with complex angles.

$$e^{\pm i (\log(|b|)+i \cdot \arg(b))}=3+b.$$

\begin{equation} \begin{cases} e^{-\arg(b)+i \cdot \log(|b|)}=3+b,\\ e^{\arg(b)-i \cdot \log(|b|)}=3+b. \end{cases} \end{equation}

Now I apply Euler's formula to get rid of the complex exponent.

\begin{equation} \begin{cases} e^{-\arg(b)}(\cos(\log(|b|)) + i \cdot \sin(\log(|b|)))=3+b,\\ e^{\arg(b)}(\cos(\log(|b|)) - i \cdot \sin(\log(|b|)))=3+b, \end{cases} \end{equation}

\begin{equation} \begin{cases} e^{-\arg(b)}\cos(\log(|b|)) + ie^{-\arg(b)} \sin(\log(|b|))=3+b,\\ e^{\arg(b)}\cos(\log(|b|)) - ie^{\arg(b)} \sin(\log(|b|))=3+b. \end{cases} \end{equation}

And now I turn the right part $b$ into polar form, so I only have $|b|$ and $\arg(b)$ in the equations.

\begin{equation} \begin{cases} e^{-\arg(b)}\cos(\log(|b|)) + ie^{-\arg(b)} \sin(\log(|b|))=3+|b|\cos(\arg(b)) + i|b|\sin(\arg(b)),\\ e^{\arg(b)}\cos(\log(|b|)) - ie^{\arg(b)} \sin(\log(|b|))=3+|b|\cos(\arg(b)) + i|b|\sin(\arg(b)), \end{cases} \end{equation}

\begin{equation} \begin{cases} e^{-\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) + i(e^{-\arg(b)} \sin(\log(|b|)) - |b|\sin(\arg(b))) = 3,\\ e^{\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) - i(e^{\arg(b)} \sin(\log(|b|)) + |b|\sin(\arg(b))) = 3. \end{cases} \end{equation}

With this, I was able to get two equation systems for each possible solution.

\begin{equation} \begin{cases} e^{-\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) = 3,\\ e^{-\arg(b)}\sin(\log(|b|)) - |b|\sin(\arg(b)) = 0, \end{cases} \end{equation}

and

\begin{equation} \begin{cases} e^{\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) = 3,\\ e^{\arg(b)}\sin(\log(|b|)) + |b|\sin(\arg(b)) = 0. \end{cases} \end{equation}

My knowledge ends here, and I have little hope of finding the solution. Is there any way to solve it other than numerically?

qianqian07
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Mowstyl
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  • The $e^{\arg(b)}\cos(\log(|b|))-|b|\cos(\arg(b))=3$ does not yield any of the real/imaginary parts of the solutions. What “previous equation” was your equation a solution to? – Тyma Gaidash Mar 20 '23 at 11:35
  • I might have made a mistake somewhere then. I'll check everything again. Thanks for pointing that out. – Mowstyl Mar 20 '23 at 11:42
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    Sorry, I forgot to answer your question. The previous equation was $f^2 = -1$, the original equation to solve was $b^f - b = 3$. – Mowstyl Mar 20 '23 at 16:30

2 Answers2

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Using Lagrange reversion on a more fun equation:

$$z^i-z+a=0\implies z=a+\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}z^{i n}}{dz^{n-1}}\right|_{z=a}$$

Use factorial power $u^{(v)}$:

$$z=\sum_{n=0}^\infty\frac{a^{(i-1)n+1}(i n)^{(n-1)}}{n!}$$

which works. The solution to $z^{-i}-z+a=0$ is the complex conjugate of the solution of $z^i-z+a=0$. Alternatively, the confluent Fox Wright function, a fairly well known special case of Fox H gives:

$$z=a\,_1\Psi_1\left(_{(2,i-1)}^{(1,i)};a^{i-1}\right)$$

Both strategies work for $a\in\Bbb C$. Maybe use $z=s^\frac1{1-i}$ to find the other root, like in Glasser’s derivation

Тyma Gaidash
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  • Thanks a lot. This actually leads to the solution. Just one thing, the factorial power (or falling factorial) should be $(in)_{(n-1)}$, what you wrote is the rising factorial. Wolfram Alpha appears to have a bug when displaying both falling and rising factorials, displaying the other one (correct me if I'm wrong, I've been using products until now instead of falling/rising factorials, they are new to me). – Mowstyl Mar 22 '23 at 11:52
  • @Mowstyl It is probably a different notation, but using $(i n)^{(n-1)}$ numerically works. The answer can be written in terms of factorials if needed? – Тyma Gaidash Mar 22 '23 at 11:55
  • No, don't worry about that. I fully understand how to obtain the solution from the Lagrange reversion theorem, thank you for that. What I was saying is that, as far as I know, the falling factorial or factorial power is written as $(a)_b$ or $a^{\underline{b}}$ and, when called in Wolfram Alpha, it is written as $(a)^b$, which is the rising factorial. It looks like a bug in Wolfram Alpha, because in the link about the factorial power you sent me they say one thing and when you execute it they do otherwise. By the way, I'm still struggling with getting the other root. Anyways, thank you a lot. – Mowstyl Mar 22 '23 at 21:50
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Just for the fun

Doing the same with any Newton-type method and searching the $ISC$, the imaginary part of $b$ as defined in the last equation is $$\Im(b)\sim \frac1{10} \,\,\Re\Big[ (-2+12 i)^{15/13}\Big]=-k$$ in an absolute error of $5.08\times 10^{-8}$.

Now, make a series expansion of $$f(a)=(a+t)^i-(a+t)-3\qquad \text{where} \qquad t=-ik$$ around $a=0$ and, truncating to some order, use power series reversion. Truncated at $O(a^{11})$, converted to decimals, this would give for this value of $k$ $$a=0.27972011$$ while the value you found is $a=0.27971992$

Claude Leibovici
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